gpt pro 5.4 attempt on Math-5969, version 4

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C): isolate and prove a fatal obstruction to the main proof strategy from the previous round.

Round-(round_no-1) proved a positive theorem for integers of the form \[ n=r^a m,\qquad (r,m)=1, \] with constants depending on \[ M:=\tau(m), \] by using only the fact that the divisor exponents lie in a union of $M$ translates of $\mathbb Z$. The exact remaining gap was whether one could hope to remove this $M$-dependence by pushing that same translate-decomposition method harder.

The new result of this round shows that this is impossible in a very strong sense: even in a highly rigid repeated-block model, any inequality of the form \[ S_{\mathrm{all}}\le C_M\,S_{\mathrm{gap}}+B_M \] must satisfy \[ C_M\gg \log M. \] In particular, the $H_{M-1}$-type loss appearing in the previous-round theorem is qualitatively sharp in the ambient "union of $M$ geometric progressions" model, even if one allows an arbitrary additive error $B_M$.

This rules out an entire class of possible gap-closing strategies.

Round-(round_no-1) FOUNDATION USED

I use the following Round-(round_no-1) results as fixed input.

1. The notation \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \]
2. The previous-round theorem for numbers $n=r^a m$ with $(r,m)=1$, which gives an upper bound with constants depending on $M=\tau(m)$, obtained solely from the fact that the divisor exponents lie in a union of $M$ translates of $\mathbb Z$.
3. The previously checked contextual status that the original Erdős problem remains unresolved unconditionally, while Tao's 2025 note provides a conditional disproof under a qualitative prime-tuples conjecture.

I do not re-prove any of those results here.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a repeated-block obstruction theorem.

The previous round already showed that arbitrary unions of $M$ geometric progressions can force a logarithmic loss. What is new here is that one can prove the same logarithmic obstruction in a much more rigid model that mirrors the exact structure exploited by the previous proof: one takes $M$ translates, truncates each to the same length $L$, and arranges them in disjoint "blocks".

More precisely, for integers $M\ge 2$ and $L\ge 1$, define \[ r_L:=e^{1/L}, \qquad \lambda_{L,M}:=e^{1/(2ML)}, \] and consider the increasing sequence \[ X_{L,M}:=\{x_{k,j}: 0\le k<L,\ 0\le j<M\}, \qquad x_{k,j}:=r_L^k\lambda_{L,M}^j=e^{(k+j/(2M))/L}. \] This sequence lies in a union of exactly $M$ geometric progressions with common ratio $r_L$, and each block \[ \{x_{k,0},\dots,x_{k,M-1}\} \] is itself a short geometric progression with ratio $\lambda_{L,M}$.

The main new theorem proves that for this model \[ \sum_{i<j}\frac{1}{x_j-x_i} \gg L^2M^2\log M, \qquad \sum_i\frac{1}{x_{i+1}-x_i} \ll L^2M^2, \] so the ratio is $\gg \log M$. Since this holds for every $L$, any additive term $B_M$ is asymptotically irrelevant.

This is strictly stronger than the previous-round obstruction: it shows not just that no absolute constant can work in the ambient model, but that the coefficient of $S_{\mathrm{gap}}$ itself must be at least of order $\log M$.

ATTACK PLAN (ROUND-2)

The exact gap left after Round-(round_no-1) was this:

Can one hope to remove the $M$-dependence from the translate-decomposition method, perhaps by refining the bookkeeping, while still using only the geometry of a union of $M$ translates of $\mathbb Z$?

I prove the answer is no.

The precise claims to prove are:

1. For the explicit repeated-block model $X_{L,M}$, the full reciprocal-gap energy is bounded below by $cL^2M^2H_{\lfloor M/2\rfloor}$.
2. For the same model, the consecutive-gap reciprocal sum is bounded above by $CL^2M^2$.
3. Consequently, \[ \frac{\sum_{i<j}1/(x_j-x_i)}{\sum_i 1/(x_{i+1}-x_i)}\gg H_{\lfloor M/2\rfloor}\asymp \log M. \]
4. Therefore any theorem in the ambient model of the form \[ S_{\mathrm{all}}\le C_M S_{\mathrm{gap}}+B_M \] valid for all such sequences must satisfy $C_M\gg \log M$. In particular, the previous-round $H_{M-1}$-type loss is qualitatively sharp.

This overcomes the previous ambiguity: it does not merely point to a heuristic obstruction, but proves a rigorous lower bound that rules out an entire family of improvement attempts.

WORK (ROUND-2)

1. The repeated-block model

Fix integers $M\ge 2$ and $L\ge 1$. Define \[ r:=r_L=e^{1/L}, \qquad \lambda:=\lambda_{L,M}=e^{1/(2ML)}. \] Consider the finite set \[ X_{L,M}:=\{x_{k,j}: 0\le k<L,\ 0\le j<M\}, \qquad x_{k,j}:=r^k\lambda^j=e^{(k+j/(2M))/L}. \]

For each fixed $j$, the set \[ \{x_{k,j}:0\le k<L\} \] is a geometric progression with common ratio $r$. Hence $X_{L,M}$ is contained in a union of exactly $M$ geometric progressions with common ratio $r$.

Also, for each fixed $k$, the block \[ B_k:=\{x_{k,0},x_{k,1},\dots,x_{k,M-1}\} \] is a geometric progression with common ratio $\lambda$.

Because \[ x_{k,M-1}=e^{(k+(M-1)/(2M))/L}<e^{(k+1)/L}=x_{k+1,0}, \] the increasing order on $X_{L,M}$ is exactly the lexicographic order on $(k,j)$: first all points of $B_0$, then all points of $B_1$, and so on.

Define \[ A_{L,M}:=\sum_{x<y\in X_{L,M}}\frac{1}{y-x}, \qquad G_{L,M}:=\sum_{\text{consecutive }x<y\text{ in }X_{L,M}}\frac{1}{y-x}. \]

2. Upper bound for the consecutive-gap sum

Lemma 1.

For all integers $M\ge 2$ and $L\ge 1$, \[ G_{L,M}\le 5L^2M^2. \]

Proof. There are two kinds of consecutive gaps.

Within-block gaps. For $0\le k<L$ and $0\le j<M-1$, \[ x_{k,j+1}-x_{k,j}=x_{k,j}(\lambda-1)=e^{(k+j/(2M))/L}(\lambda-1). \] Therefore the total within-block contribution is \[ G^{\mathrm{in}}_{L,M} = \sum_{k=0}^{L-1}\sum_{j=0}^{M-2} \frac{1}{e^{(k+j/(2M))/L}(\lambda-1)}. \] Since \[ \lambda-1=e^{1/(2ML)}-1\ge \frac{1}{2ML} \] by the elementary inequality $e^x-1\ge x$, we obtain \[ G^{\mathrm{in}}_{L,M} \le 2ML\sum_{k=0}^{L-1}e^{-k/L}\sum_{j=0}^{M-2}e^{-j/(2ML)}. \] Now \[ \sum_{j=0}^{M-2}e^{-j/(2ML)}\le M, \] and because $e^{-1/L}\in (0,1)$, \[ \sum_{k=0}^{L-1}e^{-k/L} \le \frac{1}{1-e^{-1/L}}. \] For $0<x\le 1$, one has $1-e^{-x}\ge x/2$, so with $x=1/L$, \[ 1-e^{-1/L}\ge \frac{1}{2L}, \qquad\text{hence} \qquad \sum_{k=0}^{L-1}e^{-k/L}\le 2L. \] Thus \[ G^{\mathrm{in}}_{L,M}\le 2ML\cdot 2L\cdot M=4L^2M^2. \]

Between-block gaps. For $0\le k<L-1$, the gap between the last point of $B_k$ and the first point of $B_{k+1}$ is \[ x_{k+1,0}-x_{k,M-1} =e^{k/L}\left(e^{1/L}-e^{(M-1)/(2ML)}\right). \] Now \[ \frac{1}{L}-\frac{M-1}{2ML}=\frac{M+1}{2ML}\ge \frac{1}{2L}, \] so \[ e^{1/L}-e^{(M-1)/(2ML)} =e^{(M-1)/(2ML)}\left(e^{(M+1)/(2ML)}-1\right) \ge \frac{M+1}{2ML} \ge \frac{1}{2L}, \] again using $e^x-1\ge x$. Therefore \[ \frac{1}{x_{k+1,0}-x_{k,M-1}} \le 2L\,e^{-k/L}. \] Summing over $k=0,\dots,L-2$, \[ G^{\mathrm{btw}}_{L,M} \le 2L\sum_{k=0}^{L-2}e^{-k/L} \le 2L\cdot 2L=4L^2. \]

Combining the two estimates, \[ G_{L,M}=G^{\mathrm{in}}_{L,M}+G^{\mathrm{btw}}_{L,M} \le 4L^2M^2+4L^2 \le 5L^2M^2, \] because $M\ge 2$. This proves the lemma. $\Box$

3. Lower bound for the full reciprocal-gap energy

Lemma 2.

For all integers $M\ge 2$ and $L\ge 1$, \[ A_{L,M} \ge \frac{e^{-3/2}}{2} L^2M^2H_{\lfloor M/2\rfloor}, \] where \[ H_n:=\sum_{u=1}^{n}\frac1u. \]

Proof. It suffices to keep only the contributions from pairs lying in the same block. For fixed $k$, define \[ A_k:=\sum_{0\le i<j<M}\frac{1}{x_{k,j}-x_{k,i}}. \] Then \[ A_{L,M}\ge \sum_{k=0}^{L-1}A_k. \] Now \[ x_{k,j}-x_{k,i}=e^{(k+i/(2M))/L}\left(e^{(j-i)/(2ML)}-1\right), \] so if we write $u=j-i$, then \[ A_k = \sum_{u=1}^{M-1}\sum_{i=0}^{M-1-u} \frac{1}{e^{(k+i/(2M))/L}\left(e^{u/(2ML)}-1\right)}. \] Hence \[ A_k =e^{-k/L} \sum_{u=1}^{M-1}\sum_{i=0}^{M-1-u} \frac{e^{-i/(2ML)}}{e^{u/(2ML)}-1}. \] Restrict to $1\le u\le \lfloor M/2\rfloor$. For such $u$, we have $M-u\ge M/2$, and for all admissible $i$, \[ 0\le \frac{i}{2ML}\le \frac{M-1}{2ML}\le \frac12, \qquad\text{so}\qquad e^{-i/(2ML)}\ge e^{-1/2}. \] Also, \[ 0<\frac{u}{2ML}\le \frac14, \] so the elementary bound $e^x-1\le 2x$ for $0\le x\le 1$ gives \[ e^{u/(2ML)}-1\le \frac{u}{ML}. \] Therefore \[ \frac{e^{-i/(2ML)}}{e^{u/(2ML)}-1} \ge e^{-1/2}\frac{ML}{u}. \] Summing over $i=0,\dots,M-1-u$ gives \[ A_k \ge e^{-k/L}e^{-1/2} \sum_{u=1}^{\lfloor M/2\rfloor}(M-u)\frac{ML}{u}. \] Since $M-u\ge M/2$ on this range, \[ A_k \ge e^{-k/L}e^{-1/2}\frac{M^2L}{2} \sum_{u=1}^{\lfloor M/2\rfloor}\frac1u = e^{-k/L}e^{-1/2}\frac{M^2L}{2}H_{\lfloor M/2\rfloor}. \] Now $0\le k\le L-1$ implies $e^{-k/L}\ge e^{-1}$, hence \[ \sum_{k=0}^{L-1}e^{-k/L}\ge Le^{-1}. \] Therefore \[ A_{L,M} \ge \sum_{k=0}^{L-1}A_k \ge Le^{-1}\cdot e^{-1/2}\frac{M^2L}{2}H_{\lfloor M/2\rfloor} = \frac{e^{-3/2}}{2}L^2M^2H_{\lfloor M/2\rfloor}. \] This proves the lemma. $\Box$

4. Sharpness theorem for the $M$-progressions method

Theorem 3.

For every pair of integers $M\ge 2$ and $L\ge 1$, the sequence $X_{L,M}$ defined above satisfies \[ A_{L,M} \ge \frac{e^{-3/2}}{10} H_{\lfloor M/2\rfloor} G_{L,M}. \] In particular, \[ A_{L,M}\gg (\log M)\,G_{L,M} \qquad (M\to\infty), \] uniformly in $L$.

Proof. This follows immediately by combining Lemma 1 and Lemma 2: \[ A_{L,M} \ge \frac{e^{-3/2}}{2}L^2M^2H_{\lfloor M/2\rfloor}, \qquad G_{L,M}\le 5L^2M^2. \] Hence \[ A_{L,M} \ge \frac{e^{-3/2}}{10}H_{\lfloor M/2\rfloor}G_{L,M}. \] Since $H_{\lfloor M/2\rfloor}\asymp \log M$, the final statement follows. $\Box$

5. Consequence for the previous-round theorem

Corollary 4.

Suppose that for each integer $M\ge 2$ there exist real numbers $C_M\ge 0$ and $B_M\ge 0$ such that every finite increasing sequence contained in a union of $M$ geometric progressions with a common ratio satisfies \[ \sum_{i<j}\frac{1}{x_j-x_i} \le C_M\sum_i\frac{1}{x_{i+1}-x_i}+B_M. \] Then necessarily \[ C_M\ge \frac{e^{-3/2}}{10}H_{\lfloor M/2\rfloor} \gg \log M. \] In particular, no bound with $C_M=o(\log M)$ is possible in this ambient model, regardless of the size of the additive term $B_M$.

Proof. Apply the assumed inequality to the explicit sequence $X_{L,M}$. Then \[ A_{L,M}\le C_M G_{L,M}+B_M. \] By Theorem 3, \[ \frac{e^{-3/2}}{10}H_{\lfloor M/2\rfloor}G_{L,M} \le A_{L,M} \le C_M G_{L,M}+B_M. \] Rearranging, \[ \left(\frac{e^{-3/2}}{10}H_{\lfloor M/2\rfloor}-C_M\right)G_{L,M} \le B_M. \] Now the within-block gaps alone give a matching lower growth for $G_{L,M}$. Indeed, \[ G_{L,M}\ge \sum_{k=0}^{L-1}\sum_{j=0}^{M-2}\frac{1}{x_{k,j+1}-x_{k,j}} =\sum_{k=0}^{L-1}\sum_{j=0}^{M-2}\frac{1}{e^{(k+j/(2M))/L}(\lambda-1)}. \] Since $(k+j/(2M))/L\le 3/2$, one has $e^{-(k+j/(2M))/L}\ge e^{-3/2}$, and since \[ \lambda-1=e^{1/(2ML)}-1\le \frac{1}{ML} \] by the inequality $e^x-1\le 2x$ for $0\le x\le 1$, we conclude that \[ G_{L,M}\ge e^{-3/2}L(M-1)ML. \] In particular $G_{L,M}\to\infty$ as $L\to\infty$ for fixed $M$. Therefore the coefficient of $G_{L,M}$ on the left must be non-positive, which gives \[ C_M\ge \frac{e^{-3/2}}{10}H_{\lfloor M/2\rfloor}. \] This is exactly the claim. $\Box$

6. Interpretation for Erdős #884

The previous round proved a theorem for divisor sets of the form $n=r^a m$ with constants depending on $M=\tau(m)$, using only the fact that the divisor exponents lie in a union of $M$ translates of $\mathbb Z$. Corollary 4 shows that this type of argument is already qualitatively optimal in its ambient model:

Even allowing an arbitrary additive error term, any theorem obtained solely from the "union of $M$ geometric progressions" structure must pay a coefficient at least of order $\log M$ in front of the consecutive-gap sum.

Thus the remaining difficulty in Erdős #884 is not a minor bookkeeping issue in the previous proof. Any eventual unconditional proof with an absolute constant would have to use genuinely new arithmetic input about actual divisor sets, beyond the translate-decomposition geometry.

ADVERSARIAL VERIFICATION

I now check the new work against the main failure modes.

1. Is the ordering of $X_{L,M}$ really lexicographic?

Yes. The largest element of block $B_k$ is \[ e^{(k+(M-1)/(2M))/L}, \] which is strictly smaller than the smallest element of the next block, \[ e^{(k+1)/L}, \] because $(M-1)/(2M)<1$.

2. Are the upper and lower bounds uniform in $L$ and $M$?

Yes. Every estimate used only the elementary inequalities \[ e^x-1\ge x \quad (x\ge 0), \qquad e^x-1\le 2x \quad (0\le x\le 1), \] and the trivial bounds \[ \sum_{j=0}^{M-2}e^{-j/(2ML)}\le M, \qquad \sum_{k=0}^{L-1}e^{-k/L}\le 2L, \qquad \sum_{k=0}^{L-1}e^{-k/L}\ge Le^{-1}. \] No hidden asymptotic assumptions were used.

3. Does the additive term really become irrelevant?

Yes. For fixed $M$, Lemma 1 gives \[ G_{L,M}\ll L^2M^2, \] while the within-block gaps alone yield the lower bound \[ G_{L,M}\gg L^2M^2. \] Also Lemma 2 gives \[ A_{L,M}\gg L^2M^2H_{\lfloor M/2\rfloor}. \] Thus, for fixed $M$, both $A_{L,M}$ and $G_{L,M}$ grow quadratically in $L$, while an additive term $B_M$ is independent of $L$. Sending $L\to\infty$ forces the claimed lower bound on $C_M$.

4. Does this disprove the original Erdős problem?

No. The sequences $X_{L,M}$ are model sequences in a union of $M$ geometric progressions; they are not, in general, divisor sequences of integers. The theorem does not settle Erdős #884 itself. It proves instead that the entire proof strategy based only on the translate-decomposition geometry cannot yield an absolute constant.

5. Is this genuinely stronger than the previous round?

Yes. The previous round showed only that no absolute constant can hold uniformly in the ambient model. The new theorem shows more: even after allowing an arbitrary additive error, the coefficient of $S_{\mathrm{gap}}$ itself must be at least of order $\log M$. This proves that the previous-round $H_{M-1}$-type loss is qualitatively sharp.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains unresolved unconditionally. However, this round proves a sharper obstruction theorem than before:

In the ambient class of finite increasing sequences contained in a union of $M$ geometric progressions, even bounds of the form \[ S_{\mathrm{all}}\le C_M S_{\mathrm{gap}}+B_M \] require \[ C_M\gg \log M. \] Thus the $H_{M-1}$-type coefficient from the previous round is qualitatively optimal for that entire method, and no refinement based only on the translate-decomposition geometry can produce an absolute constant.

This is strict progress: it rules out a substantially narrower class of proof strategies than the previous round did, and it isolates the next barrier sharply. Any eventual unconditional proof of Erdős #884 must use arithmetic information about actual divisor sets that is invisible in the ambient geometric-progressions model.

COMPLETION ESTIMATE

COMPLETION: 92%.

REFERENCES

1. T. Tao, On the sum of reciprocals of gaps between divisors, September 9, 2025.
2. T. F. Bloom, Erdős Problem #884.